Inequalities - GMAT Quantitative

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Question

Find the solution set of the inequality

$x^{2}+ 8x \le 4x - 4$

Tap to reveal answer

Answer

To solve a quadratic inequality, move all expressions to the left first:

$x^{2}+ 8x \le 4x - 4$

$x^{2}+ 8x - 4x + 4 \le 4x - 4 - 4x + 4$

$x^{2}+ 4x + 4 \le 0$

$(x+2) ^{2} \le 0$

The square of a real number cannot be less than 0, so

$(x+2) ^{2} = 0$

$\sqrt{(x+2) ^{2} }= \sqrt{0}$

, the only solution.

← Didn't Know|Knew It →

Question

Give the solution set of the inequality

$x^{2} +10x + 7 \le 10x + 32$

Tap to reveal answer

Answer

The square of a real number must be nonnegative, so this is a true statement regardless of the value of . The solution set is the set of all real numbers $(-\infty, \infty)$

To solve a quadratic inequality, move all expressions to the left first

$x^{2} +10x + 7 \le 10x + 32$

$x^{2} +10x + 7 - 10x - 32 \le 10x + 32 - 10x - 32$

$x^{2} -25 \le 0$

$(x+5)(x-5) \le 0$

The boundary points of the solution set will be the points at which:

; that is, ; or,

; that is,

These values will be included in the solution set, since equality is allowed by the inequality symbol.

Test the intervals

$(-\infty, -5), (-5, 5), (5, \infty)$ by choosing a value in each interval and testing the truth of the inequality.

$(-\infty, -5)$ : test

$x^{2} +10x + 7 \le 10x + 32$

$(-6) ^{2} +10 \cdot (-6) + 7 \le 10 \cdot(-6) + 32$

$36 -60 + 7 \le -60 + 32$

$-17 \le -28$

False - exclude this interval

: test

$x^{2} +10x + 7 \le 10x + 32$

$0^{2} +10 \cdot 0 + 7 \le 10 \cdot 0 + 32$

$7 \le 32$

True - include this interval

$(5, \infty)$ : test

$x^{2} +10x + 7 \le 10x + 32$

$6^{2} +10 \cdot 6 + 7 \le 10 \cdot 6 + 32$

$36 +60 + 7 \le 60 + 32$

$103 \le 92$

False - exclude this interval

is the solution set.

← Didn't Know|Knew It →

Question

Find the solution set of the inequality

$x^{2}+ 8x \le 4x - 4$

Tap to reveal answer

Answer

To solve a quadratic inequality, move all expressions to the left first:

$x^{2}+ 8x \le 4x - 4$

$x^{2}+ 8x - 4x + 4 \le 4x - 4 - 4x + 4$

$x^{2}+ 4x + 4 \le 0$

$(x+2) ^{2} \le 0$

The square of a real number cannot be less than 0, so

$(x+2) ^{2} = 0$

$\sqrt{(x+2) ^{2} }= \sqrt{0}$

, the only solution.

← Didn't Know|Knew It →

Question

Give the solution set of the inequality

$x^{2} +10x + 7 \le 10x + 32$

Tap to reveal answer

Answer

The square of a real number must be nonnegative, so this is a true statement regardless of the value of . The solution set is the set of all real numbers $(-\infty, \infty)$

To solve a quadratic inequality, move all expressions to the left first

$x^{2} +10x + 7 \le 10x + 32$

$x^{2} +10x + 7 - 10x - 32 \le 10x + 32 - 10x - 32$

$x^{2} -25 \le 0$

$(x+5)(x-5) \le 0$

The boundary points of the solution set will be the points at which:

; that is, ; or,

; that is,

These values will be included in the solution set, since equality is allowed by the inequality symbol.

Test the intervals

$(-\infty, -5), (-5, 5), (5, \infty)$ by choosing a value in each interval and testing the truth of the inequality.

$(-\infty, -5)$ : test

$x^{2} +10x + 7 \le 10x + 32$

$(-6) ^{2} +10 \cdot (-6) + 7 \le 10 \cdot(-6) + 32$

$36 -60 + 7 \le -60 + 32$

$-17 \le -28$

False - exclude this interval

: test

$x^{2} +10x + 7 \le 10x + 32$

$0^{2} +10 \cdot 0 + 7 \le 10 \cdot 0 + 32$

$7 \le 32$

True - include this interval

$(5, \infty)$ : test

$x^{2} +10x + 7 \le 10x + 32$

$6^{2} +10 \cdot 6 + 7 \le 10 \cdot 6 + 32$

$36 +60 + 7 \le 60 + 32$

$103 \le 92$

False - exclude this interval

is the solution set.

← Didn't Know|Knew It →

Question

Find the solution set of the inequality

$x^{2}+ 8x \le 4x - 4$

Tap to reveal answer

Answer

To solve a quadratic inequality, move all expressions to the left first:

$x^{2}+ 8x \le 4x - 4$

$x^{2}+ 8x - 4x + 4 \le 4x - 4 - 4x + 4$

$x^{2}+ 4x + 4 \le 0$

$(x+2) ^{2} \le 0$

The square of a real number cannot be less than 0, so

$(x+2) ^{2} = 0$

$\sqrt{(x+2) ^{2} }= \sqrt{0}$

, the only solution.

← Didn't Know|Knew It →

Question

Give the solution set of the inequality

$x^{2} +10x + 7 \le 10x + 32$

Tap to reveal answer

Answer

The square of a real number must be nonnegative, so this is a true statement regardless of the value of . The solution set is the set of all real numbers $(-\infty, \infty)$

To solve a quadratic inequality, move all expressions to the left first

$x^{2} +10x + 7 \le 10x + 32$

$x^{2} +10x + 7 - 10x - 32 \le 10x + 32 - 10x - 32$

$x^{2} -25 \le 0$

$(x+5)(x-5) \le 0$

The boundary points of the solution set will be the points at which:

; that is, ; or,

; that is,

These values will be included in the solution set, since equality is allowed by the inequality symbol.

Test the intervals

$(-\infty, -5), (-5, 5), (5, \infty)$ by choosing a value in each interval and testing the truth of the inequality.

$(-\infty, -5)$ : test

$x^{2} +10x + 7 \le 10x + 32$

$(-6) ^{2} +10 \cdot (-6) + 7 \le 10 \cdot(-6) + 32$

$36 -60 + 7 \le -60 + 32$

$-17 \le -28$

False - exclude this interval

: test

$x^{2} +10x + 7 \le 10x + 32$

$0^{2} +10 \cdot 0 + 7 \le 10 \cdot 0 + 32$

$7 \le 32$

True - include this interval

$(5, \infty)$ : test

$x^{2} +10x + 7 \le 10x + 32$

$6^{2} +10 \cdot 6 + 7 \le 10 \cdot 6 + 32$

$36 +60 + 7 \le 60 + 32$

$103 \le 92$

False - exclude this interval

is the solution set.

← Didn't Know|Knew It →

Question

Find the solution set of the inequality

$x^{2}+ 8x \le 4x - 4$

Tap to reveal answer

Answer

To solve a quadratic inequality, move all expressions to the left first:

$x^{2}+ 8x \le 4x - 4$

$x^{2}+ 8x - 4x + 4 \le 4x - 4 - 4x + 4$

$x^{2}+ 4x + 4 \le 0$

$(x+2) ^{2} \le 0$

The square of a real number cannot be less than 0, so

$(x+2) ^{2} = 0$

$\sqrt{(x+2) ^{2} }= \sqrt{0}$

, the only solution.

← Didn't Know|Knew It →

Question

Give the solution set of the inequality

$x^{2} +10x + 7 \le 10x + 32$

Tap to reveal answer

Answer

The square of a real number must be nonnegative, so this is a true statement regardless of the value of . The solution set is the set of all real numbers $(-\infty, \infty)$

To solve a quadratic inequality, move all expressions to the left first

$x^{2} +10x + 7 \le 10x + 32$

$x^{2} +10x + 7 - 10x - 32 \le 10x + 32 - 10x - 32$

$x^{2} -25 \le 0$

$(x+5)(x-5) \le 0$

The boundary points of the solution set will be the points at which:

; that is, ; or,

; that is,

These values will be included in the solution set, since equality is allowed by the inequality symbol.

Test the intervals

$(-\infty, -5), (-5, 5), (5, \infty)$ by choosing a value in each interval and testing the truth of the inequality.

$(-\infty, -5)$ : test

$x^{2} +10x + 7 \le 10x + 32$

$(-6) ^{2} +10 \cdot (-6) + 7 \le 10 \cdot(-6) + 32$

$36 -60 + 7 \le -60 + 32$

$-17 \le -28$

False - exclude this interval

: test

$x^{2} +10x + 7 \le 10x + 32$

$0^{2} +10 \cdot 0 + 7 \le 10 \cdot 0 + 32$

$7 \le 32$

True - include this interval

$(5, \infty)$ : test

$x^{2} +10x + 7 \le 10x + 32$

$6^{2} +10 \cdot 6 + 7 \le 10 \cdot 6 + 32$

$36 +60 + 7 \le 60 + 32$

$103 \le 92$

False - exclude this interval

is the solution set.

← Didn't Know|Knew It →

Question

Find the solution set of the inequality

$x^{2}+ 8x \le 4x - 4$

Tap to reveal answer

Answer

To solve a quadratic inequality, move all expressions to the left first:

$x^{2}+ 8x \le 4x - 4$

$x^{2}+ 8x - 4x + 4 \le 4x - 4 - 4x + 4$

$x^{2}+ 4x + 4 \le 0$

$(x+2) ^{2} \le 0$

The square of a real number cannot be less than 0, so

$(x+2) ^{2} = 0$

$\sqrt{(x+2) ^{2} }= \sqrt{0}$

, the only solution.

← Didn't Know|Knew It →

Question

Give the solution set of the inequality

$x^{2} +10x + 7 \le 10x + 32$

Tap to reveal answer

Answer

The square of a real number must be nonnegative, so this is a true statement regardless of the value of . The solution set is the set of all real numbers $(-\infty, \infty)$

To solve a quadratic inequality, move all expressions to the left first

$x^{2} +10x + 7 \le 10x + 32$

$x^{2} +10x + 7 - 10x - 32 \le 10x + 32 - 10x - 32$

$x^{2} -25 \le 0$

$(x+5)(x-5) \le 0$

The boundary points of the solution set will be the points at which:

; that is, ; or,

; that is,

These values will be included in the solution set, since equality is allowed by the inequality symbol.

Test the intervals

$(-\infty, -5), (-5, 5), (5, \infty)$ by choosing a value in each interval and testing the truth of the inequality.

$(-\infty, -5)$ : test

$x^{2} +10x + 7 \le 10x + 32$

$(-6) ^{2} +10 \cdot (-6) + 7 \le 10 \cdot(-6) + 32$

$36 -60 + 7 \le -60 + 32$

$-17 \le -28$

False - exclude this interval

: test

$x^{2} +10x + 7 \le 10x + 32$

$0^{2} +10 \cdot 0 + 7 \le 10 \cdot 0 + 32$

$7 \le 32$

True - include this interval

$(5, \infty)$ : test

$x^{2} +10x + 7 \le 10x + 32$

$6^{2} +10 \cdot 6 + 7 \le 10 \cdot 6 + 32$

$36 +60 + 7 \le 60 + 32$

$103 \le 92$

False - exclude this interval

is the solution set.

← Didn't Know|Knew It →

Question

Find the solution set of the inequality

$x^{2}+ 8x \le 4x - 4$

Tap to reveal answer

Answer

To solve a quadratic inequality, move all expressions to the left first:

$x^{2}+ 8x \le 4x - 4$

$x^{2}+ 8x - 4x + 4 \le 4x - 4 - 4x + 4$

$x^{2}+ 4x + 4 \le 0$

$(x+2) ^{2} \le 0$

The square of a real number cannot be less than 0, so

$(x+2) ^{2} = 0$

$\sqrt{(x+2) ^{2} }= \sqrt{0}$

, the only solution.

← Didn't Know|Knew It →

Question

Give the solution set of the inequality

$x^{2} +10x + 7 \le 10x + 32$

Tap to reveal answer

Answer

The square of a real number must be nonnegative, so this is a true statement regardless of the value of . The solution set is the set of all real numbers $(-\infty, \infty)$

To solve a quadratic inequality, move all expressions to the left first

$x^{2} +10x + 7 \le 10x + 32$

$x^{2} +10x + 7 - 10x - 32 \le 10x + 32 - 10x - 32$

$x^{2} -25 \le 0$

$(x+5)(x-5) \le 0$

The boundary points of the solution set will be the points at which:

; that is, ; or,

; that is,

These values will be included in the solution set, since equality is allowed by the inequality symbol.

Test the intervals

$(-\infty, -5), (-5, 5), (5, \infty)$ by choosing a value in each interval and testing the truth of the inequality.

$(-\infty, -5)$ : test

$x^{2} +10x + 7 \le 10x + 32$

$(-6) ^{2} +10 \cdot (-6) + 7 \le 10 \cdot(-6) + 32$

$36 -60 + 7 \le -60 + 32$

$-17 \le -28$

False - exclude this interval

: test

$x^{2} +10x + 7 \le 10x + 32$

$0^{2} +10 \cdot 0 + 7 \le 10 \cdot 0 + 32$

$7 \le 32$

True - include this interval

$(5, \infty)$ : test

$x^{2} +10x + 7 \le 10x + 32$

$6^{2} +10 \cdot 6 + 7 \le 10 \cdot 6 + 32$

$36 +60 + 7 \le 60 + 32$

$103 \le 92$

False - exclude this interval

is the solution set.

← Didn't Know|Knew It →

Question

Find the solution set of the inequality

$x^{2}+ 8x \le 4x - 4$

Tap to reveal answer

Answer

To solve a quadratic inequality, move all expressions to the left first:

$x^{2}+ 8x \le 4x - 4$

$x^{2}+ 8x - 4x + 4 \le 4x - 4 - 4x + 4$

$x^{2}+ 4x + 4 \le 0$

$(x+2) ^{2} \le 0$

The square of a real number cannot be less than 0, so

$(x+2) ^{2} = 0$

$\sqrt{(x+2) ^{2} }= \sqrt{0}$

, the only solution.

← Didn't Know|Knew It →

Question

Give the solution set of the inequality

$x^{2} +10x + 7 \le 10x + 32$

Tap to reveal answer

Answer

The square of a real number must be nonnegative, so this is a true statement regardless of the value of . The solution set is the set of all real numbers $(-\infty, \infty)$

To solve a quadratic inequality, move all expressions to the left first

$x^{2} +10x + 7 \le 10x + 32$

$x^{2} +10x + 7 - 10x - 32 \le 10x + 32 - 10x - 32$

$x^{2} -25 \le 0$

$(x+5)(x-5) \le 0$

The boundary points of the solution set will be the points at which:

; that is, ; or,

; that is,

These values will be included in the solution set, since equality is allowed by the inequality symbol.

Test the intervals

$(-\infty, -5), (-5, 5), (5, \infty)$ by choosing a value in each interval and testing the truth of the inequality.

$(-\infty, -5)$ : test

$x^{2} +10x + 7 \le 10x + 32$

$(-6) ^{2} +10 \cdot (-6) + 7 \le 10 \cdot(-6) + 32$

$36 -60 + 7 \le -60 + 32$

$-17 \le -28$

False - exclude this interval

: test

$x^{2} +10x + 7 \le 10x + 32$

$0^{2} +10 \cdot 0 + 7 \le 10 \cdot 0 + 32$

$7 \le 32$

True - include this interval

$(5, \infty)$ : test

$x^{2} +10x + 7 \le 10x + 32$

$6^{2} +10 \cdot 6 + 7 \le 10 \cdot 6 + 32$

$36 +60 + 7 \le 60 + 32$

$103 \le 92$

False - exclude this interval

is the solution set.

← Didn't Know|Knew It →

Question

Find the solution set of the inequality

$x^{2}+ 8x \le 4x - 4$

Tap to reveal answer

Answer

To solve a quadratic inequality, move all expressions to the left first:

$x^{2}+ 8x \le 4x - 4$

$x^{2}+ 8x - 4x + 4 \le 4x - 4 - 4x + 4$

$x^{2}+ 4x + 4 \le 0$

$(x+2) ^{2} \le 0$

The square of a real number cannot be less than 0, so

$(x+2) ^{2} = 0$

$\sqrt{(x+2) ^{2} }= \sqrt{0}$

, the only solution.

← Didn't Know|Knew It →

Question

Find the solution set of the inequality

$x^{2} + 6x \ge 14x - 16$

Tap to reveal answer

Answer

To solve a quadratic inequality, move all expressions to the left first:

$x^{2} + 6x \ge 14x - 16$

$x^{2} + 6x - 14x+16 \ge 14x - 16 - 14x+16$

$x^{2} -8x+16 \ge 0$

$(x-4)^{2} \ge 0$

The square of a real number must be nonnegative, so this is a true statement regardless of the value of . The solution set is the set of all real numbers $(-\infty, \infty)$

← Didn't Know|Knew It →

Question

Find the solution set of the inequality

$x^{2} + 6x \ge 14x - 16$

Tap to reveal answer

Answer

To solve a quadratic inequality, move all expressions to the left first:

$x^{2} + 6x \ge 14x - 16$

$x^{2} + 6x - 14x+16 \ge 14x - 16 - 14x+16$

$x^{2} -8x+16 \ge 0$

$(x-4)^{2} \ge 0$

The square of a real number must be nonnegative, so this is a true statement regardless of the value of . The solution set is the set of all real numbers $(-\infty, \infty)$

← Didn't Know|Knew It →

Question

Give the solution set of the inequality

$x^{2} +10x + 7 \le 10x + 32$

Tap to reveal answer

Answer

The square of a real number must be nonnegative, so this is a true statement regardless of the value of . The solution set is the set of all real numbers $(-\infty, \infty)$

To solve a quadratic inequality, move all expressions to the left first

$x^{2} +10x + 7 \le 10x + 32$

$x^{2} +10x + 7 - 10x - 32 \le 10x + 32 - 10x - 32$

$x^{2} -25 \le 0$

$(x+5)(x-5) \le 0$

The boundary points of the solution set will be the points at which:

; that is, ; or,

; that is,

These values will be included in the solution set, since equality is allowed by the inequality symbol.

Test the intervals

$(-\infty, -5), (-5, 5), (5, \infty)$ by choosing a value in each interval and testing the truth of the inequality.

$(-\infty, -5)$ : test

$x^{2} +10x + 7 \le 10x + 32$

$(-6) ^{2} +10 \cdot (-6) + 7 \le 10 \cdot(-6) + 32$

$36 -60 + 7 \le -60 + 32$

$-17 \le -28$

False - exclude this interval

: test

$x^{2} +10x + 7 \le 10x + 32$

$0^{2} +10 \cdot 0 + 7 \le 10 \cdot 0 + 32$

$7 \le 32$

True - include this interval

$(5, \infty)$ : test

$x^{2} +10x + 7 \le 10x + 32$

$6^{2} +10 \cdot 6 + 7 \le 10 \cdot 6 + 32$

$36 +60 + 7 \le 60 + 32$

$103 \le 92$

False - exclude this interval

is the solution set.

← Didn't Know|Knew It →

Question

Find the solution set of the inequality

$x^{2} + 6x \ge 14x - 16$

Tap to reveal answer

Answer

To solve a quadratic inequality, move all expressions to the left first:

$x^{2} + 6x \ge 14x - 16$

$x^{2} + 6x - 14x+16 \ge 14x - 16 - 14x+16$

$x^{2} -8x+16 \ge 0$

$(x-4)^{2} \ge 0$

The square of a real number must be nonnegative, so this is a true statement regardless of the value of . The solution set is the set of all real numbers $(-\infty, \infty)$

← Didn't Know|Knew It →

Question

Find the solution set of the inequality

$x^{2} + 6x \ge 14x - 16$

Tap to reveal answer

Answer

To solve a quadratic inequality, move all expressions to the left first:

$x^{2} + 6x \ge 14x - 16$

$x^{2} + 6x - 14x+16 \ge 14x - 16 - 14x+16$

$x^{2} -8x+16 \ge 0$

$(x-4)^{2} \ge 0$

The square of a real number must be nonnegative, so this is a true statement regardless of the value of . The solution set is the set of all real numbers $(-\infty, \infty)$

← Didn't Know|Knew It →

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